You are given integers N, K, and an integer sequence A of length M.

An integer sequence where each element is between 1 and K (inclusive) is said to be colorful when there exists a contiguous subsequence of length K of the sequence that contains one occurrence of each integer between 1 and K (inclusive).

For every colorful integer sequence of length N, count the number of the contiguous subsequences of that sequence which coincide with A, then find the sum of all the counts. Here, the answer can be extremely large, so find the sum modulo 10^9+7.

Constraints

* 1 \leq N \leq 25000
* 1 \leq K \leq 400
* 1 \leq M \leq N
* 1 \leq A_i \leq K
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N K M
A_1 A_2 ... A_M


Output

For every colorful integer sequence of length N, count the number of the contiguous subsequences of that sequence which coincide with A, then print the sum of all the counts modulo 10^9+7.

Examples

Input

3 2 1
1


Output

9


Input

4 2 2
1 2


Output

12


Input

7 4 5
1 2 3 1 2


Output

17


Input

5 4 3
1 1 1


Output

0


Input

10 3 5
1 1 2 3 3


Output

1458


Input

25000 400 4
3 7 31 127


Output

923966268


Input

9954 310 12
267 193 278 294 6 63 86 166 157 193 168 43


Output

979180369
//カラフルでない数列を数えると実装が楽
#include <iostream>
#define int long long
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

int powmod(int a, int n, int mod) {
	if (n == 0) return 1;
	if (n % 2 == 0) return powmod((a * a) % mod, n / 2, mod);
	return (a * powmod(a, n - 1, mod)) % mod;
}

bool isCourful(int a[], int M, int K) {
	int i, j;
	
	for (i = 0; i < M - K + 1; i++) {
		bool used[400] = {false};
		rep(j, K) {
			used[a[i + j]] = true;
		}
		rep(j, K) if (!used[j]) break;
		if (j == K) return true;
	}
	return false;
}

bool isPartical(int a[], int M, int K) {
	int i;
	int cnt[400] = {0};
	
	rep(i, M) {
		cnt[a[i]]++;
	}
	rep(i, K) if (cnt[i] >= 2) return false;
	return true;
}

int mod = 1000000007;
int N, K, M;
int powK[25001];
int fact[25001];
int factInv[25001];
int a[25000];
int dp1[25001][402];	//dp1[i][j] = S -> (i, j), ｛長さi, 相異なる数からなるsuffixの最大長が現在j, カラフルでない｝を全て満たす数列の個数
int dp2[25001][402];	//dp2[i][j] = (i, j) -> G,｛残り長さi, 相異なる数からなるsuffixの最大長が現在j, カラフルでない｝を全て満たす数列を1つ取ったとき、
						//カラフルになるように残りi個の要素を埋める方法の数.
						//↑これらは、「カラフル列の判定法」を変形すると導出できる。

int allSearch() {
	int i, j, k;
	static int b[25000];
	int ans = 0;
	
	rep(i, N - M + 1) {
		int all = 0;
		int courful = 0;
		rep(j, powK[N]) {
			rep(k, N) {
				b[k] = (j / powK[k]) % K;
			}
			rep(k, M) {
				if (a[k] != b[i + k]) break;
			}
			if (k < M) continue;
			all++;
			if (isCourful(b, N, K)) {
				//rep(k, N) cout << b[k] << " "; cout << endl;
				courful++;
			}
		}
		//cout << "allSearch: i = " << i << ", all = " << all;
		//cout << ", courful = " << courful << ", no = " << all - courful << endl;
		ans += courful;
		ans %= mod;
	}
	return ans;
}

int solvePartical() {
	int i, j, k;
	int ret = 0;
	
	for (i = 0; i < N - M + 1; i++) {
		int ssum = 0;
		for (j = M; j < K; j++) {
			int res = dp1[i + M][j] * dp2[N - i - M][j] % mod;
			ssum += res;
			ssum %= mod;
			//cout << "i = " << i << ", j = " << j << ", dp1[" << i + M << "][" << j << "] = " << dp1[i + M][j];
			//cout << ", dp2[" << N - i - M << "][" << j << "] = " << dp2[N - i - M][j] << ", res = " << res << endl;
		}
		int all = powK[N - M] * fact[K] % mod * factInv[K - M] % mod;
		//cout << "i = " << i << ", ssum = " << ssum << ", all = " << all << endl;
		//cout << endl;
		ret += (all - ssum + mod) % mod;
		ret %= mod;
	}
	//cout << "ret = " << ret << endl;
	ret = ret * fact[K - M] % mod * factInv[K] % mod;
	return ret;
}

int solveEasy() {
	int i;
	int pre, suf;
	bool used[400];
	
	rep(i, K) used[i] = false;
	rep(i, M) {
		if (used[a[i]]) break;
		used[a[i]] = true;
	}
	pre = i;
	
	rep(i, K) used[i] = false;
	rep(i, M) {
		if (used[a[M - 1 - i]]) break;
		used[a[M - 1 - i]] = true;
	}
	suf = i;
	
	int ret = 0;
	for (i = 0; i < N - M + 1; i++) {
		int res = dp2[i][pre] * dp2[N - M - i][suf] % mod;
		res = (powK[N - M] - res + mod) % mod;
		ret += res;
		ret %= mod;
	}
	return ret;
}

signed main() {
	int n, m, f, i;
	
	cin >> N >> K >> M;
	rep(i, M) { cin >> a[i]; a[i]--; }
	
	powK[0] = 1;
	for (i = 1; i <= N; i++) powK[i] = (powK[i - 1] * K) % mod;
	fact[0] = 1;
	factInv[0] = 1;
	for (i = 1; i <= N; i++) {
		fact[i] = (fact[i - 1] * i) % mod;
		factInv[i] = powmod(fact[i], mod - 2, mod);
	}
	
	if (isCourful(a, M, K)) {
		cout << (N - M + 1) * powK[N - M] % mod << endl;
		return 0;
	}
	
	dp1[0][0] = 1;
	rep(n, N) {
		rep(m, K) {
			dp1[n + 1][1] += dp1[n][m];
			dp1[n + 1][1] %= mod;
			dp1[n + 1][m + 1] += mod - dp1[n][m];
			dp1[n + 1][m + 1] %= mod;
		}
		
		rep(m, K) {
			dp1[n + 1][m + 1] += dp1[n + 1][m];
			dp1[n + 1][m + 1] %= mod;
		}
		
		rep(m, K) {
			if (m + 1 < K) {
				dp1[n + 1][m + 1] += dp1[n][m] * (K - m);
				dp1[n + 1][m + 1] %= mod;
			}
		}
	}
	
	rep(m, K) dp2[0][m] = 1;
	for (n = 1; n <= N; n++) {
		int rui[401] = {0};
		rep(m, K) {
			rui[m + 1] = dp2[n - 1][m];
		}
		rep(m, K) {
			rui[m + 1] += rui[m];
			rui[m + 1] %= mod;
		}
		
		rep(m, K) {
			dp2[n][m] += (rui[m + 1] - rui[1] + mod) % mod;
			dp2[n][m] %= mod;
			dp2[n][m] += dp2[n - 1][m + 1] * (K - m);
			dp2[n][m] %= mod;
		}
	}
	
	if (isPartical(a, M, K)) {
		cout << solvePartical() << endl;
	}
	else {
		cout << solveEasy() << endl;
	}
	
	//cout << "AllSearch" << endl;
	//cout << allSearch() << endl;
	return 0;
}