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>
using namespace std;
int64_t a[100405][20], off[500][20], sum[20];
int bs = 400, cm = 0;
bool check(int i, int j, int p) {
  return a[i][j] + off[i / bs][j] - a[p][j] - off[p / bs][j] == i - p;
}
int main() {
  int n, m, i, x, j;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) {
    scanf("%d", &x);
    for (j = 0; j <= 17; j++) {
      if ((1 << j) & x) {
        if (i > 0)
          a[i][j] = a[i - 1][j] + 1;
        else
          a[i][j] = 1;
      }
      sum[j] += a[i][j];
    }
  }
  while (m--) {
    int p, v;
    int64_t ans = 0;
    scanf("%d%d", &p, &v);
    p--;
    for (j = 0; j <= 17; j++) {
      int bn = p / bs;
      bool flag = a[p][j] + off[bn][j], flag2 = ((1 << j) & v);
      if (flag2 == flag) continue;
      for (i = bn * bs; off[bn][j] != 0 && i < bn * bs + bs; i++)
        a[i][j] += off[bn][j];
      off[bn][j] = 0;
      if (!flag) {
        x = 1;
        if (p > 0) x += a[p - 1][j] + off[(p - 1) / bs][j];
      } else
        x = -a[p][j];
      sum[j] += x;
      for (i = p + 1; i < n;) {
        if (!check(i, j, p)) break;
        if (i % bs == 0 && check(i + bs - 1, j, p)) {
          off[i / bs][j] += x;
          i += bs;
          sum[j] += 1LL * bs * x;
        } else {
          a[i][j] += x;
          i++;
          sum[j] += x;
        }
      }
      if (x < 0)
        a[p][j] = 0;
      else
        a[p][j] = x;
    }
    for (j = 0; j <= 17; j++) ans += (1LL << j) * sum[j];
    printf("%I64d\n", ans);
  }
}