Problem: problem

There are n cards with numbers from 1 to n. First, make a pile of cards by stacking them in order so that the top is the number 1 card, the second from the top is the number 2 card, ..., and the bottom is the number n card.

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Sort the cards by performing the following operation called "shuffle (x, y)" on the pile of cards (x, y is an integer that satisfies 1 ≤ x <y <n).

Shuffle (x, y)
n cards, pile A consisting of the xth cards from the top, pile B consisting of the x + 1th to yth cards, y + 1 consisting of the nth cards Divide into three mountains of mountain C. Then, mountain B is placed on top of mountain A, and mountain C is placed on top of it.

For example, if you perform "shuffle (3,5)" on 9 cards in order, the numbers written on the 9 cards will be 6, 7, 8, 9, 4 in order from the top. , 5, 1, 2, 3.

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Shuffle m times from the state of the first pile "Shuffle (x1, y1)" "Shuffle (x2, y2)" ... Count from the top in the pile of cards after performing "shuffle (xm, ym)" in order Create a program to find out how many cards with numbers r or less are included in the pth to qth cards.



input

The input consists of multiple datasets. Each dataset is given in the following format.

The input consists of m + 3 lines. The number of cards n is written on the first line (3 ≤ n ≤ 1000000000 = 109). The second line contains the integer m, which represents the number of shuffles (1 ≤ m ≤ 5000). The integers p, q, r are written on the third line (1 ≤ p ≤ q ≤ n, 1 ≤ r ≤ n). On the third line of i + (1 ≤ i ≤ m), two integers xi and yi (1 ≤ xi <yi <n) are written separated by a blank.

When n is 0, it indicates the end of input. The number of data sets does not exceed 5.

output

For each data set, in the pile of cards after m shuffling, output the number of cards whose number is r or less in the pth to qth cards counted from the top in one line. ..

Examples

Input

9
1
3 7 4
3 5
12
3
3 8 5
3 8
2 5
6 10
0


Output

2
3


Input

None


Output

None

Solution (in CPP):

Solution:
#define _USE_MATH_DEFINES
#include <iostream>
#include <memory>
#include <memory.h>
#include <fstream>
#include <cmath>
#include <math.h>
#include <numeric>
#include <vector>
#include <stack>
#include <string>
#include <queue>
#include <sstream>
#include <cstdlib>
#include <cassert>
#include <cstdio>
#include <map>
#include <iomanip>
#include <list>
#include <cctype>
#include <algorithm>
#include <complex>

#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, m, n) for(int i = m; i < n; i++)
using namespace std;

typedef complex<double> xy_t;
typedef pair<xy_t, xy_t> line;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<double, double> Pd;
typedef pair<int, P> PP;
typedef pair<int, PP> PPP;
const int INF = 1 << 29;
const double EPS = 1E-10;

#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, m, n) for(int i = m; i < n; i++)

int sz[10000];

void change(vector<P> &nums, int x, int y){
	vector<P> res;
	int n = nums.size();
	int indx, indy;
	int ub = n + 1, lb = 0;
	sz[0] = 0;
	rep(i, n){
		sz[i+1] = sz[i] + nums[i].second - nums[i].first + 1;
	}

	while(ub - lb > 1){
		int mb = (ub + lb) / 2;
		if(sz[mb] < x) lb = mb;
		else ub = mb;
	}
	indx = lb;
	ub = n + 1, lb = 0;

	while(ub - lb > 1){
		int mb = (ub + lb ) / 2;
		if(sz[mb] <= y) lb = mb;
		else ub = mb;
	}
	indy = lb;

	int dy = indy == n ? 0 : y - sz[indy], dx = indx == n ? 0 : x - sz[indx];

	res.push_back(P(nums[indy].first + dy ,nums[indy].second));
	rep2(i, indy + 1, n){
		res.push_back(nums[i]);
	}
	if(indx != indy) res.push_back(P(nums[indx].first + dx, nums[indx].second));
	else res.push_back(P(nums[indx].first + dx, nums[indx].first + dy-1));
	rep2(i, indx + 1, indy){
		res.push_back(nums[i]);
	}
	if(indx != indy && dy != 0) res.push_back(P(nums[indy].first, nums[indy].first + dy - 1));
	rep(i, indx ){
		res.push_back(nums[i]);
	}
	if(dx != 0)res.push_back(P(nums[indx].first, nums[indx].first + dx - 1));
	nums = res;
}

int get_num(P p, int r){
	//cout << "***" << p.first << " " << p.second << endl;;
	if(p.second < p.first) return 0;
	if(r < p.first) return 0;
	else if(p.second <= r) return p.second - p.first + 1;
	else return r - p.first + 1;
}

int counter(const vector<P> &nums, int p, int q, int r){
	int cnt = 1;
	int res = 0;
	int n = nums.size();
	rep(i, n){
	//	cout << "+ " <<  nums[i].first << " " << nums[i].second << endl;
		int sz = nums[i].second - nums[i].first + 1;
		int pn = nums[i].first + p - cnt;
		int qn = nums[i].first + q - cnt;
		
		res += get_num(P(max(nums[i].first, pn), min(nums[i].second, qn)), r);
		cnt += sz;
	}
	return res;
}

int main(){
	int n, m;
	int p, q, r;
	int x, y;
	while(cin >> n && n){
		vector<P> nums;
		nums.push_back(P(1, n));
		cin >> m;
		cin >> p >> q >> r;
		rep(i, m){
			cin >> x >> y;
		    change(nums, x, y);
		}
		cout << counter(nums, p, q, r) << endl;
	}
	return 0;
}