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 arr[100001], pre[21][4 * 100001], suff[21][4 * 100001],
    ans[21][4 * 100001];
inline int check(long long int val, long long int bit) {
  if (val & (1LL << bit)) return 1;
  return 0;
}
void build(long long int id, long long int node, long long int a,
           long long int b) {
  if (a == b) {
    ans[id][node] = check(arr[a], id);
    pre[id][node] = check(arr[a], id);
    suff[id][node] = check(arr[a], id);
    return;
  }
  long long int mid = (a + b) / 2;
  build(id, 2 * node, a, mid);
  build(id, 2 * node + 1, mid + 1, b);
  pre[id][node] = pre[id][2 * node];
  if (pre[id][2 * node] == mid - a + 1) pre[id][node] += pre[id][2 * node + 1];
  suff[id][node] = suff[id][2 * node + 1];
  if (suff[id][2 * node + 1] == b - (mid + 1) + 1)
    suff[id][node] += suff[id][2 * node];
  ans[id][node] = ans[id][2 * node] + ans[id][2 * node + 1];
  ans[id][node] += suff[id][2 * node] * pre[id][2 * node + 1];
}
void update(long long int id, long long int node, long long int a,
            long long int b, long long int ind, long long int val) {
  if (a > b || a > ind || b < ind) return;
  if (a == b) {
    ans[id][node] = check(val, id);
    pre[id][node] = check(val, id);
    suff[id][node] = check(val, id);
    return;
  }
  long long int mid = (a + b) / 2;
  if (ind <= mid)
    update(id, 2 * node, a, mid, ind, val);
  else
    update(id, 2 * node + 1, mid + 1, b, ind, val);
  pre[id][node] = pre[id][2 * node];
  if (pre[id][2 * node] == mid - a + 1) pre[id][node] += pre[id][2 * node + 1];
  suff[id][node] = suff[id][2 * node + 1];
  if (suff[id][2 * node + 1] == b - (mid + 1) + 1)
    suff[id][node] += suff[id][2 * node];
  ans[id][node] = ans[id][2 * node] + ans[id][2 * node + 1];
  ans[id][node] += suff[id][2 * node] * pre[id][2 * node + 1];
}
int main() {
  ios::sync_with_stdio(false);
  ;
  cin.tie(0);
  ;
  long long int i, j, n, q, ind, val, res = 0;
  cin >> n >> q;
  for (i = 1; i <= n; i++) cin >> arr[i];
  for (i = 0; i < 20; i++) build(i, 1, 1, n);
  while (q--) {
    cin >> ind >> val;
    for (i = 0; i < 20; i++) update(i, 1, 1, n, ind, val);
    res = 0;
    for (i = 0; i < 20; i++) res += (1LL << i) * ans[i][1] * 1LL;
    cout << res << "\n";
  }
  return 0;
}