1420_D. Rescue Nibel!
Ori and Sein have overcome many difficult challenges. They finally lit the Shrouded Lantern and found Gumon Seal, the key to the Forlorn Ruins. When they tried to open the door to the ruins... nothing happened.

Ori was very surprised, but Sein gave the explanation quickly: clever Gumon decided to make an additional defence for the door.

There are n lamps with Spirit Tree's light. Sein knows the time of turning on and off for the i-th lamp — l_i and r_i respectively. To open the door you have to choose k lamps in such a way that there will be a moment of time when they all will be turned on.

While Sein decides which of the k lamps to pick, Ori is interested: how many ways there are to pick such k lamps that the door will open? It may happen that Sein may be wrong and there are no such k lamps. The answer might be large, so print it modulo 998 244 353.

Input

First line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n) — total number of lamps and the number of lamps that must be turned on simultaneously.

Next n lines contain two integers l_i ans r_i (1 ≤ l_i ≤ r_i ≤ 10^9) — period of time when i-th lamp is turned on.

Output

Print one integer — the answer to the task modulo 998 244 353.

Examples

Input


7 3
1 7
3 8
4 5
6 7
1 3
5 10
8 9


Output


9

Input


3 1
1 1
2 2
3 3


Output


3

Input


3 2
1 1
2 2
3 3


Output


0

Input


3 3
1 3
2 3
3 3


Output


1

Input


5 2
1 3
2 4
3 5
4 6
5 7


Output


7

Note

In first test case there are nine sets of k lamps: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 6), (1, 4, 6), (2, 3, 6), (2, 4, 6), (2, 6, 7).

In second test case k=1, so the answer is 3.

In third test case there are no such pairs of lamps.

In forth test case all lamps are turned on in a time 3, so the answer is 1.

In fifth test case there are seven sets of k lamps: (1, 2), (1, 3), (2, 3), (2, 4), (3, 4), (3, 5), (4, 5).
```python
import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c for j in range(b)] for i in range(a)]
def list3d(a, b, c, d): return [[[d for k in range(c)] for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e for l in range(d)] for k in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10**19
MOD = 998244353
EPS = 10**-10

def compress(S):

    zipped, unzipped = {}, {}
    for i, a in enumerate(sorted(S)):
        zipped[a] = i
        unzipped[i] = a
    return zipped, unzipped

class ModTools:

    def __init__(self, MAX, MOD):

        MAX += 1
        self.MAX = MAX
        self.MOD = MOD
        factorial = [1] * MAX
        factorial[0] = factorial[1] = 1
        for i in range(2, MAX):
            factorial[i] = factorial[i-1] * i % MOD
        inverse = [1] * MAX
        inverse[MAX-1] = pow(factorial[MAX-1], MOD-2, MOD)
        for i in range(MAX-2, -1, -1):
            inverse[i] = inverse[i+1] * (i+1) % MOD
        self.fact = factorial
        self.inv = inverse

    def nCr(self, n, r):

        if n < r: return 0
        r = min(r, n-r)
        numerator = self.fact[n]
        denominator = self.inv[r] * self.inv[n-r] % self.MOD
        return numerator * denominator % self.MOD

N, K = MAP()
LR = []
S = set()
for i in range(N):
    l, r = MAP()
    r += 1
    LR.append((l, r))
    S.add(l)
    S.add(r)

zipped, _ = compress(S)
M = len(zipped)
lcnt = [0] * M
rcnt = [0] * M
for i in range(N):
    LR[i] = (zipped[LR[i][0]], zipped[LR[i][1]])
    lcnt[LR[i][0]] += 1
    rcnt[LR[i][1]] += 1

cur = 0
ans = 0
mt = ModTools(N, MOD)
for i in range(M):
    cur -= rcnt[i]
    while lcnt[i]:
        if cur >= K-1:
            ans += mt.nCr(cur, K-1)
            ans %= MOD
        cur += 1
        lcnt[i] -= 1
print(ans)

```