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 maxn = 100000 + 5;
int n, m;
int a[maxn];
set<pair<int, int> > seg[32];
long long add(set<pair<int, int> > &sg, int pos, int k) {
  pair<int, int> now = make_pair(pos, n + 1);
  set<pair<int, int> >::iterator it = sg.lower_bound(now);
  int cnt_prev = 0, cnt_next = 0, nl = pos, nr = pos;
  if (it != sg.begin()) {
    it--;
    if (it->second == pos - 1) {
      cnt_prev = it->second - it->first + 1;
      nl = it->first;
      sg.erase(it);
    }
  }
  it = sg.lower_bound(now);
  if (it != sg.end()) {
    if (it->first == pos + 1) {
      cnt_next = it->second - it->first + 1;
      nr = it->second;
      sg.erase(it);
    }
  }
  sg.insert(make_pair(nl, nr));
  return (1LL << k) * (cnt_prev + 1) * (cnt_next + 1);
}
long long del(set<pair<int, int> > &sg, int pos, int k) {
  pair<int, int> now = make_pair(pos, n + 1);
  set<pair<int, int> >::iterator it = sg.upper_bound(now);
  it--;
  int cnt_prev = pos - it->first, cnt_next = it->second - pos;
  if (pos != it->first) sg.insert(make_pair(it->first, pos - 1));
  if (pos != it->second) sg.insert(make_pair(pos + 1, it->second));
  sg.erase(it);
  return (1LL << k) * (cnt_prev + 1) * (cnt_next + 1) * -1;
}
void printseg(set<pair<int, int> > &sg) {
  set<pair<int, int> >::iterator it = sg.begin();
  while (it != sg.end()) {
    printf("[%d,%d] ", it->first, it->second);
    it++;
  }
  puts("");
}
long long update(int p, int v) {
  long long ret = 0;
  for (int i = 0; i < 32; i++) {
    int bita = (bool)(a[p] & (1 << i)), bitb = (bool)(v & (1 << i));
    if (bita == 0 && bitb == 1) ret += add(seg[i], p, i);
    if (bita == 1 && bitb == 0) ret += del(seg[i], p, i);
  }
  a[p] = v;
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    sum += update(i, tmp);
  }
  for (int i = 1; i <= m; i++) {
    int p, v;
    scanf("%d%d", &p, &v);
    sum += update(p, v);
    cout << sum << endl;
  }
  return 0;
}