There are $N$ robots who work for $Y$ days and on each day they
produce some toys .on some days a few robots are given rest.
So depending on the availability of robots owner has 
made a time table which decides which robots will work on 
the particular day.
Only contiguous robots must be selected as they can form 
a link of communication among themselves.
Initially, all robots have the capacity of one toy.
On each day capacity for the chosen robot is updated i.e
capacity = capacity $+$$ ($minimum capacity of given range % $1000000007)$ .
After calculating the minimum capacity of a given range, compute
it as modulo 1000000007 ($10^9 + 7$).
After $Y$ days find the minimum capacity of the $N$ robots and compute it as modulo 1000000007 ($10^9 + 7$).   

-----Input:-----
- First-line will contain $T$, the number of test cases. Then the test cases follow. 
- Next Line contains a single integer N.
- Next Line contains a single integer Y.
- Next $Y$ lines contains l and r range of chosen robots .

-----Output:-----
For each testcase, output in a single line answer , the minimum capacity of the $N$ robots after $Y$ days and compute it as modulo 1000000007 ($10^9 + 7$) .

-----Constraints-----
- $1 \leq T \leq 100$
- $100 \leq N \leq 10^4$
- $200 \leq Y \leq 1000$
- $0<=l , r<=N-1$  , $l<=r$

-----Sample Input:-----
1
5
4
0 3
1 2
4 4
0 4

-----Sample Output:-----
4

-----EXPLANATION:-----
Initial capacity of the $5$ robots
1 1 1 1 1 
Minimum in range [0,3] = 1
Update the capacity in the range [0,3]
Now capacity becomes, 
Day 1 - 2 2 2 2 1
Similarly capacities changes for each day 
Day 2 - 2 4 4 2 1
Day 3 - 2 4 4 2 2
Day 4 - 4 6 6 4 4
so after 4 days minimum capacity is $4$.

Source: codechef | Difficulty: UNKNOWN_DIFFICULTY


Solutions:
MAX = 100005
tree = [0] * MAX
lazy = [0] * MAX

def updateRangeUtil(si, ss, se, us, ue, diff):
	if lazy[si] != 0:
		tree[si] += lazy[si]
		if ss != se:
			lazy[si * 2 + 1] += lazy[si]
			lazy[si * 2 + 2] += lazy[si]
		lazy[si] = 0
	if ss > se or ss > ue or se < us:
		return
	if ss >= us and se <= ue:
		tree[si] += diff
		if ss != se:
			lazy[si * 2 + 1] += diff
			lazy[si * 2 + 2] += diff
		return
	mid = (ss + se) // 2
	updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff)
	updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff)
	tree[si] = min(tree[si * 2 + 1], tree[si * 2 + 2])

def updateRange(n, us, ue, diff):
	updateRangeUtil(0, 0, n - 1, us, ue, diff)

def getSumUtil(ss, se, qs, qe, si):
	if lazy[si] != 0:
		tree[si] += lazy[si]
		if ss != se:
			lazy[si * 2 + 1] += lazy[si]
			lazy[si * 2 + 2] += lazy[si]
		lazy[si] = 0
	if ss > se or ss > qe or se < qs:
		return 10000000000.0
	if ss >= qs and se <= qe:
		return tree[si]
	mid = (ss + se) // 2
	return min(getSumUtil(ss, mid, qs, qe, 2 * si + 1), getSumUtil(mid + 1, se, qs, qe, 2 * si + 2))

def getSum(n, qs, qe):
	if qs < 0 or qe > n - 1 or qs > qe:
		return -1
	return getSumUtil(0, n - 1, qs, qe, 0)

def constructSTUtil(arr, ss, se, si):
	if ss > se:
		return
	if ss == se:
		tree[si] = arr[ss]
		return
	mid = (ss + se) // 2
	constructSTUtil(arr, ss, mid, si * 2 + 1)
	constructSTUtil(arr, mid + 1, se, si * 2 + 2)
	tree[si] = min(tree[si * 2 + 1], tree[si * 2 + 2])

def constructST(arr, n):
	constructSTUtil(arr, 0, n - 1, 0)
for _ in range(int(input())):
	tree = [0] * MAX
	lazy = [0] * MAX
	n = int(input())
	y = int(input())
	arr = [1] * n
	constructST(arr, n)
	for xyz in range(y):
		(l, r) = list(map(int, input().split()))
		updateRange(n, l, r, getSum(n, l, r) % 1000000007)
	print(getSum(n, 0, n - 1) % 1000000007)