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 double EPS = 1e-9;
const double PI = acos(-1);
const int INF = (int)1e9 + 9;
const int MAXN = (int)1e5 + 7;
int n, m;
int a[20][MAXN];
long long ans[20];
set<pair<int, int> > second[20];
long long first(int x) { return 1ll * x * (x + 1) / 2; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    for (int bit = 0; bit < 20; bit++) a[bit][i] = (x >> bit) & 1;
  }
  for (int bit = 0; bit < 20; bit++) {
    int i = 1;
    while (i <= n) {
      if (a[bit][i]) {
        int j = i;
        while (j <= n && a[bit][j]) j++;
        second[bit].insert(make_pair(i, j - 1));
        ans[bit] += first(j - i);
        i = j;
      } else {
        i++;
      }
    }
  }
  while (m--) {
    int pos, val;
    scanf("%d%d", &pos, &val);
    for (int bit = 0; bit < 20; bit++) {
      int cur = (val >> bit) & 1;
      if (a[bit][pos] == cur) continue;
      if (cur) {
        int l = pos, r = pos;
        auto nxt = second[bit].upper_bound(make_pair(pos, INF));
        if (nxt != second[bit].end()) {
          if (nxt->first == pos + 1) {
            ans[bit] -= first(nxt->second - nxt->first + 1);
            r = nxt->second;
            second[bit].erase(nxt);
          }
        }
        auto prv = second[bit].upper_bound(make_pair(pos, INF));
        if (prv != second[bit].begin()) {
          prv--;
          if (prv->second == pos - 1) {
            ans[bit] -= first(prv->second - prv->first + 1);
            l = prv->first;
            second[bit].erase(prv);
          }
        }
        ans[bit] += first(r - l + 1);
        second[bit].insert(make_pair(l, r));
      } else {
        auto it = second[bit].upper_bound(make_pair(pos, INF));
        assert(it != second[bit].begin());
        it--;
        int l = it->first;
        int r = it->second;
        ans[bit] -= first(r - l + 1);
        second[bit].erase(it);
        if (l < pos) {
          ans[bit] += first(pos - l);
          second[bit].insert(make_pair(l, pos - 1));
        }
        if (r > pos) {
          ans[bit] += first(r - pos);
          second[bit].insert(make_pair(pos + 1, r));
        }
      }
      a[bit][pos] = cur;
    }
    long long res = 0;
    for (int i = 0; i < 20; i++) res += (1LL << i) * ans[i];
    printf("%lld\n", res);
  }
  return 0;
}