C. Star sky 二维前缀和

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si(1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

Examples

input

2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5

output

3
0
3

input

3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51

output

3
3
5
0

Note

Let's consider the first example.

At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.

At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.

At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
typedef long long LL;
#define MAXN 109
#define N 100
/*
坐标的最大值比较小 开前缀数组求和
*/
LL a[MAXN][MAXN][];
LL n, q, c;
int main()
{
    cin >> n >> q >> c;
    int t1, t2, t3, t4;
    for (LL i = ; i < n; i++)
    {
        cin >> t1 >> t2 >> t3;
        a[t1][t2][t3]++;
    }
    for(int i =;i<=;i++)
        for(int j=;j<=;j++)
            for (int k = ; k < c + ; k++)
            {
                a[i][j][k] += a[i - ][j][k] + a[i][j - ][k] - a[i - ][j - ][k];
             }
    int T;
    int x1, y1, x2, y2;
    while (q--)
    {
        cin >> T >> x1 >> y1 >> x2 >> y2;
        LL cnt = , sum = ;
        for (int i = ; i < c + ; i++)
        {
            cnt = a[x2][y2][i] - a[x2][y1 - ][i] - a[x1 - ][y2][i] + a[x1 - ][y1 - ][i];
            sum += cnt*((i + T) % (c + ));
        }
        cout << sum << endl;
    }
}

错误的二分代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
typedef long long LL;
#define MAXN 100005
#define N 100
/*
二分查找区域内符合范围的点，取模运算求和
...
*/
struct star
{
    LL x, y, s;
    bool operator<(const star& rhs)
    {
        if (x == rhs.x)
            return y < rhs.y;
        else
            return x < rhs.x;
    }
}a[MAXN];
LL n, q, c;
LL t, X1, Y1, X2, Y2;
LL sum = ;
void solve(LL l, LL r)
{
    if (l > r)
        return;
    if (l == r)
    {
        if (a[l].x >= X1&&a[l].x >= Y1&&a[l].x <= X2&&a[l].y <= Y2)
            sum += (a[l].s + t) % (c + );
        return;
    }
    LL mid = (l + r) / ;
    //cout << mid << endl;
    if (a[mid].x > X2)
        solve(l, mid - );
    else if (a[mid].x < X1)
        solve(mid + , r);
    else
    {
        if (a[mid].x >= X1&&a[mid].x >= Y1&&a[mid].x <= X2&&a[mid].y <= Y2)
            sum += (a[mid].s + t) % (c + );
        solve(l, mid - );
        solve(mid + , r);
    }
}
int main()
{
    scanf("%lld%lld%lld", &n, &q, &c);
    for (LL i = ; i < n; i++)
        scanf("%lld%lld%lld", &a[i].x, &a[i].y, &a[i].s);
    sort(a, a + n);
    while (q--)
    {
        sum = ;
        scanf("%lld%lld%lld%lld%lld", &t, &X1, &Y1, &X2, &Y2);
        solve(, n-);
        printf("%lld\n", sum);
    }
    return ;
}