Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

3 4
1 1 1
1 1
2 2
3 2
1 2


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int maxn = 100050, maxb = 20;
int n, m, d = 17, A[maxn];
long long ans = 0;
int main() {
  int d0, i, j, p, v;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (d0 = 0; d0 < d; d0++)
    for (i = 1; i <= n; i++)
      if ((A[i] >> d0) & 1) {
        for (j = i + 1; j <= n && (A[j] >> d0) & 1; j++)
          ;
        ans += ((long long)(j - i) * (j - i + 1) / 2) << d0;
        i = j;
      }
  for (i = 0; i < m; i++) {
    scanf("%d%d", &p, &v);
    for (d0 = 0; d0 < d; d0++)
      if ((v >> d0 & 1) ^ (A[p] >> d0 & 1)) {
        int ls, rs;
        for (j = p + 1; j <= n && A[j] >> d0 & 1; j++)
          ;
        ls = j - p;
        for (j = p - 1; j && A[j] >> d0 & 1; j--)
          ;
        rs = p - j;
        if (A[p] >> d0 & 1)
          ans -= ((long long)ls * rs) << d0;
        else
          ans += ((long long)ls * rs) << d0;
      }
    A[p] = v;
    printf("%I64d\n", ans);
  }
  return 0;
}