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;
int n, m;
long long ans = 0;
long long ff(long long x) { return x * (x + 1) / 2; }
int a[100010] = {};
int p, v;
long long t[100010] = {};
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  t[0] = 1;
  for (int i = 1; i <= 19; i++) t[i] = t[i - 1] * 2;
  for (int i = 0; i <= 19; i++) {
    for (int j = 1; j <= n; j++) {
      if ((a[j] & (1 << i)))
        for (int k = j; k <= n; k++) {
          if ((a[k + 1] & (1 << i)) == 0) {
            ans += ff(k - j + 1) * t[i];
            j = k;
            break;
          }
        }
    }
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &p, &v);
    for (int j = 0; j <= 19; j++) {
      if ((v & (1 << j)) != (a[p] & (1 << j))) {
        long long z = 0;
        long long y = 0;
        for (int k = p - 1; k >= 1; k--) {
          if (a[k] & (1 << j)) {
            z++;
          } else
            break;
        }
        for (int k = p + 1; k <= n; k++) {
          if (a[k] & (1 << j)) {
            y++;
          } else
            break;
        }
        long long o = ff(z + y + 1) - ff(z) - ff(y);
        if (v & (1 << j))
          ans += o * t[j];
        else
          ans -= o * t[j];
      }
    }
    a[p] = v;
    printf("%I64d\n", ans);
  }
  return 0;
}