#include <bits/stdc++.h>
using namespace std;

const int N = 200005;
int T[N << 2], lz[N << 2];

#define mid ((l + r) >> 1)
void push(int v, int l, int r) {
	if (lz[v] == 0) return;
	if (l < r) lz[v << 1] += lz[v], lz[v << 1 | 1] += lz[v];
	T[v] += lz[v]; T[v] = min(T[v], N); lz[v] = 0; 
}

void upd_pos(int v, int l, int r, int x, int val) {
	push(v, l, r);
	if (l > r || l > x || r < x) return;
	if (l == r) { T[v] = val; return; }
	upd_pos(v << 1, l, mid, x, val); upd_pos(v << 1 | 1, mid + 1, r, x, val);
	T[v] = min(T[v << 1], T[v << 1 | 1]);
}

void upd(int v, int l, int r, int L, int R, int val) {
	push(v, l, r);
	if (l > r || R < l || L > r) return;
	if (L <= l && r <= R) { lz[v] = val; push(v, l, r); return; }
	upd(v << 1, l, mid, L, R, val); upd(v << 1 | 1, mid + 1, r, L, R, val);
	T[v] = min(T[v << 1], T[v << 1 | 1]);
}

int get(int v, int l, int r, int L, int R) {
	push(v, l, r);
	if (l > r || R < l || L > r) return N;
	if (L <= l && r <= R) return T[v];
	return min(get(v << 1, l, mid, L, R), get(v << 1 | 1, mid + 1, r, L, R));
}
#undef mid

int n, q;
int b[N];
vector<int> Q[N];
int f[N][2];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &b[i]);
		if (b[i] == 0) f[i][1] = 1; else f[i][0] = 1;
	}
	scanf("%d", &q);
	while(q--) {
		int l, r; scanf("%d %d", &l, &r);
		Q[l].push_back(r);
	}

	for (int i = 0; i < 4 * N; ++i) T[i] = N;
	upd_pos(1, 0, n, 0, 0);

	for (int i = 1; i <= n; ++i) {
		sort(Q[i].begin(), Q[i].end(), greater<int>());
		for (int j : Q[i]) {
			int mn = get(1, 0, n, 0, j);
			upd_pos(1, 0, n, j, mn);
		}
		upd(1, 0, n, 0, i - 1, f[i][0]);
		upd(1, 0, n, i, n, f[i][1]);
	}

	printf("%d\n", get(1, 0, n, 0, n));
}
### Prompt

Create a solution in CPP for the following problem:

You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

const int N = 200005;
int T[N << 2], lz[N << 2];

#define mid ((l + r) >> 1)
void push(int v, int l, int r) {
	if (lz[v] == 0) return;
	if (l < r) lz[v << 1] += lz[v], lz[v << 1 | 1] += lz[v];
	T[v] += lz[v]; T[v] = min(T[v], N); lz[v] = 0; 
}

void upd_pos(int v, int l, int r, int x, int val) {
	push(v, l, r);
	if (l > r || l > x || r < x) return;
	if (l == r) { T[v] = val; return; }
	upd_pos(v << 1, l, mid, x, val); upd_pos(v << 1 | 1, mid + 1, r, x, val);
	T[v] = min(T[v << 1], T[v << 1 | 1]);
}

void upd(int v, int l, int r, int L, int R, int val) {
	push(v, l, r);
	if (l > r || R < l || L > r) return;
	if (L <= l && r <= R) { lz[v] = val; push(v, l, r); return; }
	upd(v << 1, l, mid, L, R, val); upd(v << 1 | 1, mid + 1, r, L, R, val);
	T[v] = min(T[v << 1], T[v << 1 | 1]);
}

int get(int v, int l, int r, int L, int R) {
	push(v, l, r);
	if (l > r || R < l || L > r) return N;
	if (L <= l && r <= R) return T[v];
	return min(get(v << 1, l, mid, L, R), get(v << 1 | 1, mid + 1, r, L, R));
}
#undef mid

int n, q;
int b[N];
vector<int> Q[N];
int f[N][2];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &b[i]);
		if (b[i] == 0) f[i][1] = 1; else f[i][0] = 1;
	}
	scanf("%d", &q);
	while(q--) {
		int l, r; scanf("%d %d", &l, &r);
		Q[l].push_back(r);
	}

	for (int i = 0; i < 4 * N; ++i) T[i] = N;
	upd_pos(1, 0, n, 0, 0);

	for (int i = 1; i <= n; ++i) {
		sort(Q[i].begin(), Q[i].end(), greater<int>());
		for (int j : Q[i]) {
			int mn = get(1, 0, n, 0, j);
			upd_pos(1, 0, n, j, mn);
		}
		upd(1, 0, n, 0, i - 1, f[i][0]);
		upd(1, 0, n, i, n, f[i][1]);
	}

	printf("%d\n", get(1, 0, n, 0, n));
}
```