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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 5;
int dp[N], b[N], cnt[N], n, q, st[4 * N];
vector <int> endPoints[N];

int Left(int x) {return (x << 1);}
int Right(int x) {return (x << 1) + 1;}

void update(int id, int L, int R, int i, int j, int val) {
	if (L > j || R < i) return;
	if (L >= i && R <= j) {
		st[id] = min(st[id], val);
		return;
	}
	int mid = (L + R) >> 1;
	update(Left(id), L, mid, i, j, val);
	update(Right(id), mid + 1, R, i, j, val);
}

int get(int id, int L, int R, int i) {
	if (i == 0) return 0;
	if (L > i || R < i) return N;
	if (L == R) return st[id];
	int mid = (L + R) >> 1;
	if (i <= mid) return min(st[id], get(Left(id), L, mid, i));
	else return min(st[id], get(Right(id), mid + 1, R, i));
}

int main(int argc, char const *argv[])
{
	ios_base::sync_with_stdio(0); cin.tie(0);

	cin >> n;
	for (int i = 1; i <= 4 * n; ++i)
		st[i] = N;
	for (int i = 1; i <= n; ++i) {
		cin >> b[i];
		cnt[i] = cnt[i - 1] + (b[i] == 0 ? 1 : 0);
	}
	cin >> q;
	while (q--) {
		int l, r;
		cin >> l >> r;
		endPoints[r].push_back(l);
	}
	for (int i = 1; i <= n; ++i) {
		update(1, 1, n, i, i, get(1, 1, n, i - 1) + (b[i] == 0 ? -1 : 1));
		for (int l : endPoints[i]) {
			update(1, 1, n, l, i, get(1, 1, n, l - 1));
		}
	}
	cout << cnt[n] + get(1, 1, n, n) << endl;

	return 0;
}