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>
int FASTBUFFER;
using namespace std;
int n, m, i, j, x, now, p, v, a[100005];
set<int> s[25];
set<int>::iterator it;
long long ans;
int main() {
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 0; i <= 20; i++) {
    s[i].insert(n + 1);
    x = (1 << i);
    now = n + 1;
    for (j = n; j >= 1; j--) {
      if ((a[j] & x) == 0)
        s[i].insert(j), now = j;
      else
        ans += (long long)(now - j) * x;
    }
  }
  while (m--) {
    scanf("%d %d", &p, &v);
    for (i = 0; i <= 20; i++) {
      x = (1 << i);
      if ((v & x) == (a[p] & x)) continue;
      if ((v & x) == 0) {
        int next, succ;
        it = s[i].upper_bound(p);
        next = *it;
        if (it == s[i].begin())
          succ = 0;
        else
          it--, succ = *it;
        ans -= (long long)(p - succ) * (long long)(next - p) * x;
        s[i].insert(p);
      } else {
        int next, succ;
        it = s[i].upper_bound(p);
        next = *it;
        it = s[i].lower_bound(p);
        if (it == s[i].begin())
          succ = 0;
        else
          it--, succ = *it;
        ans += (long long)(p - succ) * (long long)(next - p) * x;
        s[i].erase(p);
      }
    }
    cout << ans << endl;
    a[p] = v;
  }
  return 0;
}