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 n, m, k, i, j, a, l, r, g, u, v, st, st1, st2;
vector<long long> f[20];
long long A[20][100010];
long long ans[20], res;
void upd(int wh, int x, int d) {
  for (int i = x; i < n; i = (i | (i + 1))) f[wh][i] += d;
}
long long sum(int wh, int x) {
  int ret = 0;
  for (int i = x; i >= 0; i = (i & (i + 1)) - 1) ret += f[wh][i];
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (k = 1; k <= 18; k++)
    for (i = 0; i < n; i++) f[k].push_back(0);
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    k = 0;
    while (a > 0) {
      k++;
      if (a % 2 == 1) {
        upd(k, i, 1);
        A[k][i] = 1;
      }
      a /= 2;
    }
  }
  for (k = 1; k <= 18; k++)
    for (i = 0; i < n; i++)
      if (A[k][i] == 1) {
        j = i;
        while (j < n && A[k][j] == 1) j++;
        ans[k] += ((j - i) * (j - i + 1)) / 2;
        i = j;
      }
  while (m--) {
    scanf("%d%d", &v, &u);
    v--;
    k = 0;
    while (k < 18) {
      k++;
      if (A[k][v] == u % 2) {
        u /= 2;
        continue;
      }
      if (u % 2 == 0) {
        r = v;
        l = 0;
        st = sum(k, v);
        while (l + 1 < r) {
          g = (l + r) / 2;
          if (sum(k, g - 1) + v - g + 1 == st)
            r = g;
          else
            l = g + 1;
        }
        if (sum(k, l - 1) + v - l + 1 == st)
          st1 = l;
        else
          st1 = r;
        l = v;
        r = n - 1;
        st = sum(k, v);
        while (l + 1 < r) {
          g = (l + r) / 2;
          if (sum(k, g) - (g - v) == st)
            l = g;
          else
            r = g - 1;
        }
        if (sum(k, r) - r + v == st)
          st2 = r;
        else
          st2 = l;
        ans[k] =
            ans[k] + ((v - st1) * (v - st1 + 1) + (st2 - v) * (st2 - v + 1) -
                      (st2 - st1 + 1) * (st2 - st1 + 2)) /
                         2;
        A[k][v] = 0;
        upd(k, v, -1);
      } else {
        A[k][v] = 1;
        upd(k, v, 1);
        r = v;
        l = 0;
        st = sum(k, v);
        while (l + 1 < r) {
          g = (l + r) / 2;
          if (sum(k, g - 1) + v - g + 1 == st)
            r = g;
          else
            l = g + 1;
        }
        if (sum(k, l - 1) + v - l + 1 == st)
          st1 = l;
        else
          st1 = r;
        l = v;
        r = n - 1;
        st = sum(k, v);
        while (l + 1 < r) {
          g = (l + r) / 2;
          if (sum(k, g) + v - g == st)
            l = g;
          else
            r = g - 1;
        }
        if (sum(k, r) + v - r == st)
          st2 = r;
        else
          st2 = l;
        ans[k] =
            ans[k] + ((st2 - st1 + 1) * (st2 - st1 + 2) -
                      (v - st1) * (v - st1 + 1) - (st2 - v) * (st2 - v + 1)) /
                         2;
      }
      u /= 2;
    }
    res = 0;
    for (long long k = 1; k <= 18; k++)
      res += (long long)(ans[k] * (1 << (k - 1)));
    cout << res << endl;
  }
}