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;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
const int oo = 1e9;
const int MOD = 1000000007;
const int BASE = 26;
const int N = 202010;
const int MX = 200000;
#define x first
#define y second
#define pb push_back
#define mp make_pair

int n , m , b[N], seg[4 * N], lazy[4 * N], dp[N], gg;
pii iv[N];
vector<int> g[N];


void build(int n , int s , int e) {
	lazy[n] = oo;
	if (s == e) {
		seg[n] = oo;
		return;
	}
	build(2 * n, s, (s + e) / 2);
	build(2 * n + 1, (s + e) / 2 + 1, e);
	seg[n] = min(seg[2 * n], seg[2 * n + 1]);
}
void push(int n , int s , int e) {
	seg[n] = min(seg[n], lazy[n]);
	if (s != e) {
		lazy[2 * n] = min(lazy[2 * n], lazy[n]);
		lazy[2 * n + 1] = min(lazy[2 * n + 1], lazy[n]);
	}
	lazy[n] = oo;
}
void update(int n, int s , int e, int l ,  int r , int val) {
	push(n, s, e);
	if (s > r || e < l)return;
	if (s >= l && e <= r) {
		lazy[n] = val;
		push(n, s, e);
		return;
	}
	update(2 * n, s, (s + e) / 2, l, r, val);
	update(2 * n + 1, (s + e) / 2 + 1, e, l, r, val);
	seg[n] = min(seg[2 * n], seg[2 * n + 1]);
}
int get(int n , int s , int e, int l , int r) {
	push(n, s, e);
	if (s > r || e < l)return oo;
	if (s >= l && e <= r)
		return seg[n];
	return min(get(2 * n, s, (s + e) / 2, l, r), get(2 * n + 1, (s + e) / 2 + 1, e, l, r));
}
int main() {
	scanf("%d", &n);
	for (int i = 1 ; i <= n; ++i) {
		scanf("%d", &b[i]);
		if (!b[i])++gg;
	}
	++n;
	build(1, 1, n);
	update(1, 1, n, n, n, 0);
	scanf("%d", &m);
	for (int i = 1; i <= m; ++i) {
		scanf("%d%d", &iv[i].x, &iv[i].y);
		g[iv[i].x].pb(iv[i].y);
	}
	for (int i = n - 1 ; i >= 1 ; --i) {
		dp[i] = (b[i] ? 1 : -1) + dp[i + 1];
		update(1, 1, n, i, i, dp[i]);
		for (auto j : g[i]) {
			dp[i] = min(dp[i], get(1, 1, n, j + 1, j + 1));
			update(1, 1, n, i, j, get(1, 1, n, j + 1, j + 1));
		}
	}


	printf("%d\n", dp[1] + gg);
	return 0;
}