Given are integers N and K, and a prime number P. Find the number, modulo P, of directed graphs G with N vertices that satisfy below. Here, the vertices are distinguishable from each other.

* G is a tournament, that is, G contains no duplicated edges or self-loops, and exactly one of the edges u\to v and v\to u exists for any two vertices u and v.
* The in-degree of every vertex in G is at most K.
* For any four distinct vertices a, b, c, and d in G, it is not the case that all of the six edges a\to b, b\to c, c\to a, a\to d, b\to d, and c\to d exist simultaneously.

Constraints

* 4 \leq N \leq 200
* \frac{N-1}{2} \leq K \leq N-1
* 10^8<P<10^9
* N and K are integers.
* P is a prime number.

Input

Input is given from Standard Input in the following format:


N K P


Output

Print the number of directed graphs that satisfy the conditions, modulo P.

Examples

Input

4 3 998244353


Output

56


Input

7 3 998244353


Output

720


Input

50 37 998244353


Output

495799508
#include <bits/stdc++.h>

#define debug(...) fprintf(stderr, __VA_ARGS__)

#ifndef AT_HOME
#define getchar() IO::myGetchar()
#define putchar(x) IO::myPutchar(x)
#endif

namespace IO {
	static const int IN_BUF = 1 << 23, OUT_BUF = 1 << 23;

	inline char myGetchar() {
		static char buf[IN_BUF], *ps = buf, *pt = buf;
		if (ps == pt) {
			ps = buf, pt = buf + fread(buf, 1, IN_BUF, stdin);
		}
		return ps == pt ? EOF : *ps++;
	}

	template<typename T>
	inline bool read(T &x) {
		bool op = 0;
		char ch = getchar();
		x = 0;
		for (; !isdigit(ch) && ch != EOF; ch = getchar()) {
			op ^= (ch == '-');
		}
		if (ch == EOF) {
			return false;
		}
		for (; isdigit(ch); ch = getchar()) {
			x = x * 10 + (ch ^ '0');
		}
		if (op) {
			x = -x;
		}
		return true;
	}

	inline int readStr(char *s) {
		int n = 0;
		char ch = getchar();
		for (; isspace(ch) && ch != EOF; ch = getchar())
			;
		for (; !isspace(ch) && ch != EOF; ch = getchar()) {
			s[n++] = ch;
		}
		s[n] = '\0';
		return n;
	}

	inline void myPutchar(char x) {
		static char pbuf[OUT_BUF], *pp = pbuf;
		struct _flusher {
			~_flusher() {
				fwrite(pbuf, 1, pp - pbuf, stdout);
			}
		};
		static _flusher outputFlusher;
		if (pp == pbuf + OUT_BUF) {
			fwrite(pbuf, 1, OUT_BUF, stdout);
			pp = pbuf;
		}
		*pp++ = x;
	}

	template<typename T>
	inline void print_(T x) {
		if (x == 0) {
			putchar('0');
			return;
		}
		static int num[40];
		if (x < 0) {
			putchar('-');
			x = -x;
		}
		for (*num = 0; x; x /= 10) {
			num[++*num] = x % 10;
		}
		while (*num){
			putchar(num[*num] ^ '0');
			--*num;
		}
	}

	template<typename T>
	inline void print(T x, char ch = '\n') {
		print_(x);
		putchar(ch);
	}

	inline void printStr_(const char *s, int n = -1) {
		if (n == -1) {
			n = strlen(s);
		}
		for (int i = 0; i < n; ++i) {
			putchar(s[i]);
		}
	}

	inline void printStr(const char *s, int n = -1, char ch = '\n') {
		printStr_(s, n);
		putchar(ch);
	}
}
using namespace IO;

const int N = 205;

int n, k, P, fac[N], inv[N];

int qpow(int a, int b) {
	int s = 1;
	for (; b; b >>= 1) {
		if (b & 1) {
			s = 1ll * s * a % P;
		}
		a = 1ll * a * a % P;
	}
	return s;
}

void init(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; ++i) {
		fac[i] = 1ll * fac[i - 1] * i % P;
	}
	inv[n] = qpow(fac[n], P - 2);
	for (int i = n; i; --i) {
		inv[i - 1] = 1ll * inv[i] * i % P;
	}
}

int f[N][N];

int calc(int k, int l, int m) {
	for (int i = 1; i <= k; ++i) {
		int sum = 0;
		for (int j = l; j >= 0; --j) {
			sum = i > 1 ? (sum + f[i - 1][j]) % P : 1;
			if (i - 1 + j <= m && k - i + l - j <= m) {
				f[i][j] = sum;
			} else {
				f[i][j] = 0;
			}
		}
		f[1][0] = 0;
	}
	return f[k][0];
}

int solve(int n, int m) {
	if (n == 1) {
		return 1;
	}
	int ans = 0;
	for (int i = 2; i < n; ++i) {
		ans = (ans + calc(i, n - i, m)) % P;
	}
	return 1ll * ans * fac[n - 1] % P;
}

int main() {
	read(n), read(k), read(P);
	init(n);
	int ans = 0;
	for (int i = 0; i <= k; ++i) {
		ans = (ans + 1ll * fac[n] * inv[n - i] % P * solve(n - i, k - i)) % P;
	}
	// debug("%d\n", calc(4, 3, 3));
	print(ans);
}
