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;
int a[100010];
long long ans = 0;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int Q = 0; Q < 20; Q++) {
    for (int j = 1; j <= n; j++)
      if (a[j] & (1 << Q)) {
        int k = 0;
        for (k = j; k <= n && (a[k] & (1 << Q)); k++)
          ;
        k--;
        ans += ((long long)(k - j + 1) * (k - j + 2) / 2) << Q;
        j = k;
      }
  }
  for (int P = 1; P <= m; P++) {
    int p, v;
    scanf("%d%d", &p, &v);
    for (int Q = 0; Q < 20; Q++)
      if ((a[p] & (1 << Q)) ^ (v & (1 << Q))) {
        int l, r;
        for (l = p - 1; l && (a[l] & (1 << Q)); l--)
          ;
        for (r = p + 1; r <= n && (a[r] & (1 << Q)); r++)
          ;
        l = p - l;
        r -= p;
        if ((v >> Q) & 1)
          ans += ((long long)l * r) << Q;
        else
          ans -= ((long long)l * r) << Q;
      }
    a[p] = v;
    printf("%I64d\n", ans);
  }
  return 0;
}