Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:

* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.



Note that the initial values of ai (i = 1, 2, . . . , n) are 0.

Constraints

* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.

Input


n q
com1 x1 y1
com2 x2 y2
...
comq xq yq


In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).

Output

For each getSum operation, print the sum in a line.

Example

Input

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


Output

3
2
#include <bits/stdc++.h>

#define DEBUG(C) cerr << #C << " = " << C << endl

using namespace std;

template<typename T>
class BIT {
private:
  vector<T> bit;
public:
  BIT(const int n) : bit(n + 10) {}
  void add(int idx, T val) {
    ++idx;
    while (idx <= bit.size()) {
      bit[idx] += val;
      idx += idx & -idx;
    }
  }
  T get(int l, int r) {
    return this->get(r) - this->get(l - 1);
  }
  T get(int r) {
    ++r;
    T res = 0;
    while (r > 0) {
      res += bit[r];
      r -= r & -r;
    }
    return res;
  }
};

int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  BIT<int> bit(n);
  while (q--) {
    int c, x, y;
    scanf("%d%d%d", &c, &x, &y);
    if (c) {
      printf("%d\n", bit.get(x - 1, y - 1));
    } else {
      bit.add(x - 1, y);
    }
  }
  return 0;
}