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;
long long int bt[30][100011];
long long int num[30][100011];
void update(long long int *a, long long int val, long long int ind) {
  while (ind <= 100011 && ind > 0) {
    a[ind] += val;
    ind += (ind & -ind);
  }
}
long long int query(long long int *a, long long int ind) {
  long long int sum = 0;
  while (ind > 0) {
    sum += a[ind];
    ind -= (ind & -ind);
  }
  return sum;
}
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int a[n + 2];
  a[0] = 0;
  a[n + 2] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    long long int t = a[i];
    long long int cur = 0;
    while (t > 0) {
      num[cur][i] = t % 2;
      t /= 2;
      cur++;
    }
    for (int j = 0; j < 29; j++) {
      if (num[j][i]) update(bt[j], 1, i);
    }
  }
  long long int ans = 0;
  for (int j = 0; j < 29; j++) {
    long long int t = 0;
    for (int i = 1; i <= n; i++) {
      if (num[j][i]) {
        ans = ans + (t + 1) * (1LL << j);
        t++;
      } else
        t = 0;
    }
  }
  long long int p, v;
  while (m--) {
    cin >> p >> v;
    for (int j = 0; j < 29; j++) {
      if (((1LL << j) & v) == ((1LL << j) & a[p])) continue;
      long long int n1 = 0, n2 = 0;
      if (num[j][p - 1] == 0) {
        n1 = 0;
      } else {
        long long int hi = p - 1;
        long long int lo = 1;
        long long int mid;
        while (hi > lo) {
          mid = (hi + lo) / 2;
          if (query(bt[j], p - 1) - query(bt[j], mid - 1) == p - mid) {
            hi = mid;
          } else
            lo = mid + 1;
        }
        n1 = p - hi;
      }
      if (num[j][p + 1] == 0) {
        n2 = 0;
      } else {
        long long int hi = n;
        long long int lo = p + 1;
        long long int mid;
        while (hi > lo) {
          mid = (hi + lo) / 2 + 1;
          if (query(bt[j], mid) - query(bt[j], p) == mid - p) {
            lo = mid;
          } else
            hi = mid - 1;
        }
        n2 = lo - p;
      }
      if ((a[p] & (1LL << j)) != 0) {
        update(bt[j], -1, p);
        ans = ans - (n1 + n2 + n1 * n2 + 1) * (1LL << j);
      } else {
        update(bt[j], 1, p);
        ans = ans + (n1 + n2 + n1 * n2 + 1) * (1LL << j);
      }
    }
    a[p] = v;
    for (int j = 0; j < 29; j++) {
      num[j][p] = v % 2;
      v /= 2;
    }
    cout << ans << "\n";
  }
}