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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl


const int inf = 1e9;
template<typename T> struct segtree {
    
    T merge(T x, T y) {
	return min(x,y);
    }
    int n;
    vector<int> t;
    void init(int n) {
	n += 10;
	this->n = n;
	t.resize(n*4);
    }
 
    void build(int v, int tl, int tr) {
	if (tl==tr) {
	    t[v] = inf;
	} else {
	    int tm = (tl+tr)/2;
	    build(2*v,tl,tm);
	    build(2*v+1,tm+1,tr);
	    t[v] = merge(t[2*v], t[2*v+1]);
	}
    }
 
    void upd(int v, int tl, int tr, int i, T dx) {
	if (tl == tr) {
	    t[v] = merge(t[v], dx);
	} else {
	    int tm = (tl+tr)/2;
	    if (i<=tm) {
		upd(2*v,tl,tm,i,dx);
	    } else {
		upd(2*v+1,tm+1,tr,i,dx);
	    }
	    t[v] = merge(t[2*v], t[2*v+1]);
	}
    }
 
    T qry(int v, int tl, int tr, int l, int r) {
	if (l>tr || r<tl) {
	    return inf;
	}
	if (l <= tl && tr <= r) {
	    return t[v];
	}
	int tm = (tl+tr)/2;
	return merge(qry(2*v,tl,tm,l,r), qry(2*v+1,tm+1,tr,l,r));
    }
};



const int maxn = 1e6 + 5;


int n;
int b[maxn];
vector<int> iv[maxn];

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

    cin>>n;
    int b0=0;
    
    for (int i=0; i<n; i++) {
	cin>>b[i];
	if (b[i]==0) {
	    b0++;
	    b[i]=-1;
	}
    }

    segtree<int> t;
    t.init(n+10);
    t.build(1,1,n+1);


    int q; cin>>q;
    while (q--) {
	int l,r;
	cin>>l>>r;
	--l;
	// [l,r)
	iv[l].push_back(r);
    }

    // dp[i]: min penalty considering first i terms of b.
    auto qry = [&](int l, int r) {
	return t.qry(1,1,n+1,l+1,r+1);
    };


    auto upd = [&](int i, int val) {
	t.upd(1,1,n+1,i+1,val);
    };

    for (int i=0; i<=n; i++) {
	if (i>0) {
	    upd(i, qry(i-1,i-1) + b[i-1]);
	} else {
	    upd(i, 0);
	}

	for (int r: iv[i]) {
	    upd(r, qry(i,r-1));
	}
    }

    int res = b0+qry(n,n);
    cout<<res<<endl;
    
    
    return 0;
}