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;
const int N = 1e5 + 7;
const int SS = 4 * N;
const int LG = 20;
long long seg[SS][LG], suf[SS][LG], pre[SS][LG];
int n, m;
void update(int ind, int bit, bool val, int s = 0, int e = n, int v = 1) {
  if (e - s == 1) {
    seg[v][bit] = pre[v][bit] = suf[v][bit] = val;
    return;
  }
  int mid = (s + e) / 2;
  if (ind < mid)
    update(ind, bit, val, s, mid, 2 * v);
  else
    update(ind, bit, val, mid, e, 2 * v + 1);
  pre[v][bit] = pre[2 * v][bit];
  if (pre[2 * v][bit] == mid - s) pre[v][bit] += pre[2 * v + 1][bit];
  suf[v][bit] = suf[2 * v + 1][bit];
  if (suf[2 * v + 1][bit] == e - mid) suf[v][bit] += suf[2 * v][bit];
  seg[v][bit] = seg[2 * v][bit] + seg[2 * v + 1][bit] +
                suf[2 * v][bit] * pre[2 * v + 1][bit];
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    for (int j = 0; j < LG; j++) update(i, j, (t >> j) & 1);
  }
  while (m--) {
    int p, v;
    cin >> p >> v;
    p--;
    long long ans = 0;
    for (int i = 0; i < LG; i++) {
      update(p, i, (v >> i) & 1);
      ans += seg[1][i] * (1 << i);
    }
    cout << ans << "\n";
  }
  return 0;
}