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>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
int t[20][100010], a[20][100010];
int n, q, x[100010], y, z, bp[20];
long long f[20];
long long all;
void upd(int id, int r, int val) {
  for (int i = r; i < n; i |= i + 1) t[id][i] += val;
}
int get(int id, int r) {
  int ans = 0;
  for (int i = r; i >= 0; i = (i & (i + 1)) - 1) ans += t[id][i];
  return ans;
}
int get(int id, int l, int r) { return get(id, r) - get(id, l - 1); }
void build(int id) {
  for (int i = 0; i < n; ++i) upd(id, i, a[id][i]);
}
long long Find(int bit) {
  long long ans = 0;
  for (int i = 0, j; i < n;) {
    for (j = i + 1; j < n && a[bit][i] == a[bit][j]; ++j)
      ;
    if (a[bit][i] == 1) ans += 1ll * (j - i) * (j - i + 1) / 2;
    i = j;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
    for (int j = 0; j < 17; ++j) a[j][i] = ((x[i] & (1LL << j)) != 0);
  }
  for (int j = 0; j < 20; ++j) build(j);
  bp[0] = 1;
  for (int i = 1; i < 20; ++i) bp[i] = 2ll * bp[i - 1];
  for (int i = 0; i < 20; ++i) f[i] = Find(i);
  while (q--) {
    int y, z;
    cin >> y >> z;
    --y;
    int L, R, l, r, m;
    for (int i = 0; i < 17; ++i) {
      int Cur = ((z & (1 << i)) != 0);
      if (get(i, y) - get(i, y - 1) == Cur) continue;
      if (y - 1 >= 0) {
        int l = 0, r = y - 1;
        while (l < r - 1) {
          int m = (l + r) >> 1;
          if (get(i, m, y - 1) == y - m)
            r = m;
          else
            l = m;
        }
        if (get(i, l, y - 1) == y - l)
          L = l;
        else if (get(i, r, y - 1) == y - r)
          L = r;
        else
          L = y;
      } else
        L = y;
      if (y + 1 < n) {
        l = y + 1, r = n - 1;
        while (l < r - 1) {
          int m = (l + r) >> 1;
          if (get(i, y + 1, m) == m - (y + 1) + 1)
            l = m;
          else
            r = m;
        }
        if (get(i, y + 1, r) == r - (y + 1) + 1)
          R = r;
        else if (get(i, y + 1, l) == l - (y + 1) + 1)
          R = l;
        else
          R = y;
      } else
        R = y;
      if (Cur == 1)
        f[i] += 1ll * (R - L + 1) * (R - L + 2) / 2 -
                1ll * (y - L) * (y - L + 1) / 2 -
                1ll * (R - y) * (R - y + 1) / 2;
      else
        f[i] -= 1ll * (R - L + 1) * (R - L + 2) / 2 -
                1ll * (y - L) * (y - L + 1) / 2 -
                1ll * (R - y) * (R - y + 1) / 2;
      upd(i, y, (Cur == 1 ? 1 : -1));
    }
    all = 0;
    for (int i = 0; i < 17; ++i) all += 1ll * f[i] * bp[i];
    cout << all << endl;
  }
}