Codeforces 750E New Year and Old Subsequence - 线段树 - 动态规划

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

input

8 3
20166766
1 8
1 7
2 8

output

4
3
-1

input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15

output

-1
2
1
-1
-1

input

4 2
1234
2 4
1 2

output

-1
-1

Note

In the first sample:

• In the first query, ugliness("20166766") = 4 because all four sixes must be removed.
• In the second query, ugliness("2016676") = 3 because all three sixes must be removed.
• In the third query, ugliness("0166766") =  - 1 because it's impossible to remove some digits to get a nice string.

In the second sample:

• In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice.
• In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170".

题目大意

　　给定一个长度为$n$的数字串。每次询问一个区间至少要删除多少个数字使得包含子序列"2017"但不包含子序列"2016"，无解输出-1。

　　dp是显然的。

　　因为每次询问一个区间，所以需要把dp状态扔到某个数据结构上。先考虑线段树。

　　线段树更新的时候是拿两段的信息合并，所以不能像做1~n的dp那样记录状态。

　　考虑2017之间的间隔：

| 2 | 0 | 1 | 7 |

0   1   2   3   4

　　线段树的每个节点存一个矩阵$A$。$a_{ij}$表示使原串的子序列包含2017中第$i$个间隔到第$j$个间隔组成的子串，但不包含严格包含它的子序列最少需要删除的数字、

　　转移是显然的，和区间dp一样。枚举区间，枚举中间点，然后转移就好了。

　　考虑初值问题，显然的是非2、0、1、7、6的数字对答案不影响，所以令$a_{ii} = 0$，$a_{ij} = \infty \ \ \  \left ( i \neq j \right )$。

　　考虑当前数字是2的时候，如果我希望只包含子串$[0, 0]$（这里表示两个间隔间的子串），那么就必须删掉这个2，故$a_{00} = 1$，如果希望包含子串$[0, 1]$，那么什么都不用做，所以$a_{01} = 0$。对于0、1、7同理。

　　考虑当前数字是6的时候，那么遇到子串$[i, 3]$希望转移回自己，那么需要付出1的代价，因为否则会包含子序列"2016"，同样如果遇到子串$[i, 4]$希望转移回自己，那么也需要付出1的代价。

　　由于很早以前过的这道题，所以不想重写一份，代码有点丑，请谅解。

Code

 /**
  * Codeforces
  * Problem#750E
  * Accepted
  * Time: 998ms
  * Memory: 49276k
  */
 #include<iostream>
 #include<fstream>
 #include<sstream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<ctime>
 #include<cctype>
 #include<cmath>
 #include<algorithm>
 #include<stack>
 #include<queue>
 #include<set>
 #include<map>
 #include<vector>
 #include<malloc.h>
 #ifndef WIN32
 #define AUTO "%lld"
 #else
 #define AUTO "%I64d"
 #endif
 using namespace std;
 typedef bool boolean;
 #define inf 0xfffffff
 #define smin(a, b)    (a) = min((a), (b))
 #define smax(a, b)    (a) = max((a), (b))
 template<typename T>
 inline void readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && x != -);
     if(x == -)    return;
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
 }

 typedef class Matrix {
     public:
         int mat[][];    //0:        1:2            2:20        3:201        4:2017
         Matrix(){        }
         Matrix(char x){
             for(int i = ; i <= ; i++)
                 for(int j = ; j <= ; j++)
                     mat[i][j] = (i == j) ? () : (inf);
             if(x == '')    mat[][] = , mat[][] = ;
             else if(x == '')    mat[][] = , mat[][] = ;
             else if(x == '')    mat[][] = , mat[][] = ;
             else if(x == '')    mat[][] = , mat[][] = ;
             else if(x == '')    mat[][] = , mat[][] = ;
         }

         Matrix operator +(Matrix x) {
             Matrix res;
             for(int i = ; i < ; i++)
                 for(int j = ; j < ; j++){
                     res.mat[i][j] = inf;
                     for(int k = ; k < ; k++)
                         smin(res.mat[i][j], mat[i][k] + x.mat[k][j]);
                 }
             return res;
         }
 }Matrix;

 typedef class SegTreeNode {
     public:
         Matrix a;
         SegTreeNode* left, *right;
         SegTreeNode():left(NULL), right(NULL){        }
         SegTreeNode(char x):left(NULL), right(NULL){
             a = Matrix(x);
         }

         void pushUp() {
             a = left->a + right->a;
         }
 }SegTreeNode;

 typedef class SegTree {
     public:
         SegTreeNode* root;
         SegTree():root(NULL){        }
         SegTree(int size, char* str){
             build(root, , size, str);
         }

         void build(SegTreeNode*& node, int l, int r, char* list){
             if(l == r){
                 node = new SegTreeNode(list[l]);
                 return;
             }
             node = new SegTreeNode();
             int mid = (l + r) >> ;
             build(node->left, l, mid, list);
             build(node->right, mid + , r, list);
             node->pushUp();
         }

         Matrix query(SegTreeNode*& node, int l, int r, int from, int end) {
             if(l == from && r == end)    return node->a;
             int mid = (l + r) >> ;
             if(end <= mid)    return query(node->left, l, mid, from, end);
             if(from > mid)    return query(node->right, mid + , r, from, end);
             return query(node->left, l, mid, from, mid) + query(node->right, mid + , r, mid + , end);
         }
 }SegTree;

 int n, q;
 char* str;
 SegTree st;

 inline void init() {
     readInteger(n);
     readInteger(q);
     str = new char[(const int)(n + )];
     scanf("%s", str + );
     st = SegTree(n, str);
 }

 inline void solve() {
     int a, b;
     while(q--) {
         readInteger(a);
         readInteger(b);
         Matrix c = st.query(st.root, , n, a, b);
         printf("%d\n", (c.mat[][] == inf) ? (-) : (c.mat[][]));
     }
 }

 int main() {
     init();
     solve();
     return ;
 }