Prefix Sum
题号:NC17685
时间限制:C/C++/Rust/Pascal 3秒,其他语言6秒
空间限制:C/C++/Rust/Pascal 256 M,其他语言512 M
64bit IO Format: %lld

题目描述

Niuniu has learned prefix sum and he found an interesting about prefix sum.

Let's consider (k+1) arrays a[i] (0 <= i <= k)
The index of a[i] starts from 1. 
a[i] is always the prefix sum of a[i-1]. 
"always" means a[i] will change when a[i-1] changes.
"prefix sum" means a[i][1] = a[i-1][1] and a[i][j] = a[i][j-1] + a[i-1][j] (j >= 2)


Initially, all elements in a[0] are 0.
There are two kinds of operations, which are modify and query.
For a modify operation, two integers x, y are given, and it means a[0][x] += y.
For a query operation, one integer x is given, and it means querying a[k][x].

As the result might be very large, you should output the result mod 1000000007.

输入描述:

The first line contains three integers, n, m, k.
n is the length of each array.
m is the number of operations.
k is the number of prefix sum.

In the following m lines, each line contains an operation.

If the first number is 0, then this is a change operation.
There will be two integers x, y after 0, which means a[0][x] += y;
If the first number is 1, then this is a query operation.

There will be one integer x after 1, which means querying a[k][x].

1 <= n <= 100000
1 <= m <= 100000
1 <= k <= 40
1 <= x <= n
0 <= y < 1000000007

输出描述:

For each query, you should output an integer, which is the result.
示例1

输入

复制
4 11 3
0 1 1
0 3 1
1 1
1 2
1 3
1 4
0 3 1
1 1
1 2
1 3
1 4

输出

复制
1
3
7
13
1
3
8
16

说明

For the first 4 queries, the (k+1) arrays are
1 0 1 0
1 1 2 2
1 2 4 6
1 3 7 13
For the last 4 queries, the (k+1) arrays are
1 0 2 0
1 1 3 3
1 2 5 8
1 3 8 16