Star sky
Contest: Codeforces Round 427 (Div. 2)
Difficulty rating: 1600
Tags: dp, implementation

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

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

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.

Example 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

Example output:
3
0
3

Example 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

Example output:
3
3
5
0

Editorial:
The brightness of the i-th star in moment t is $$( s _ { i } + t _ { i } ) \bmod ( c + 1 )$$, where $$\bmod$$ is modulo operator.
Let's precalc for each p = 0...С, x = 1...100, y = 1...100 cnt[p][x][y] — the number of stars with the initial brightness p in the rectangle (1; 1)-(x; y). It is calculated like calcuating of partial sums: cnt[p][x][y] = cnt[p][x - 1][y] + cnt[p][x][y - 1] - cnt[p][x - 1][y - 1] + (the number of stars in the point (x;y) with the initial brightness p).
Then the total brightness of stars at the i-th view is $$\sum_{p=0}^{c} \left( (p + t_i) \mod (c+1) \right) * \text{amount}(p, x_{1i}, y_{1i}, x_{2i}, y_{2i})$$, where $$amount(p,x_{1},y_{1},x_{2},y_{2})$$ is the number of stars with the initial brightness p in the given rectangle. This number can be calculated using array $$\text{cnt}$$, using the inclusion-exclusion principle: amount(p, x1, y1, x2, y2) = cnt[p][x2][y2] - cnt[p][x1 - 1][y2] - cnt[p][x2][y1 - 1] + cnt[p][x1 - 1][y1 - 1].
Let X is the maximum coordinate. Time complexity: O(n + qc + cX2). Memory complexity: O(cX2).