#include <bits/stdc++.h>
using namespace std;
long long powmod(long long a, long long b, long long MOD) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
const int N = 1e6 + 5;
const int MOD = 1E9 + 7;
int a[N];
int b[N];
int dis[N];
int dp[2][N];
inline void addmod(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int getsum(int lim, int pre) {
  int sum = 0;
  for (int i = 0; i < lim; i++) {
    addmod(sum, dp[pre][i]);
  }
  return sum;
}
int getsum2(int sum, long long l, int k, int n) {
  l = l / n - k + 1;
  return l < 1 ? 0 : l % MOD * sum % MOD;
}
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    dis[i] = a[i];
  }
  sort(dis, dis + n);
  int m = unique(dis, dis + n) - dis;
  int cur = 1, pre = 0, add = 0, lim = l % n;
  for (int i = 0; i < n; i++) {
    dp[pre][i] = 1;
    a[i] = lower_bound(dis, dis + m, a[i]) - dis;
  }
  for (int i = 1; i <= k && (i - 1) * n < l; i++) {
    for (int j = 0; j < m; j++) {
      b[j] = 0;
    }
    for (int j = 0; j < n; j++) {
      addmod(b[a[j]], dp[pre][j]);
    }
    for (int j = 1; j < m; j++) {
      addmod(b[j], b[j - 1]);
    }
    int sum = 0;
    for (int j = 0; j < n; j++) {
      addmod(sum, dp[pre][j]);
    }
    addmod(add, getsum(lim, pre));
    addmod(add, getsum2(sum, l, i, n));
    for (int j = 0; j < n; j++) {
      dp[cur][j] = b[a[j]];
    }
    swap(cur, pre);
  }
  printf("%d\n", add);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long powmod(long long a, long long b, long long MOD) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
const int N = 1e6 + 5;
const int MOD = 1E9 + 7;
int a[N];
int b[N];
int dis[N];
int dp[2][N];
inline void addmod(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int getsum(int lim, int pre) {
  int sum = 0;
  for (int i = 0; i < lim; i++) {
    addmod(sum, dp[pre][i]);
  }
  return sum;
}
int getsum2(int sum, long long l, int k, int n) {
  l = l / n - k + 1;
  return l < 1 ? 0 : l % MOD * sum % MOD;
}
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    dis[i] = a[i];
  }
  sort(dis, dis + n);
  int m = unique(dis, dis + n) - dis;
  int cur = 1, pre = 0, add = 0, lim = l % n;
  for (int i = 0; i < n; i++) {
    dp[pre][i] = 1;
    a[i] = lower_bound(dis, dis + m, a[i]) - dis;
  }
  for (int i = 1; i <= k && (i - 1) * n < l; i++) {
    for (int j = 0; j < m; j++) {
      b[j] = 0;
    }
    for (int j = 0; j < n; j++) {
      addmod(b[a[j]], dp[pre][j]);
    }
    for (int j = 1; j < m; j++) {
      addmod(b[j], b[j - 1]);
    }
    int sum = 0;
    for (int j = 0; j < n; j++) {
      addmod(sum, dp[pre][j]);
    }
    addmod(add, getsum(lim, pre));
    addmod(add, getsum2(sum, l, i, n));
    for (int j = 0; j < n; j++) {
      dp[cur][j] = b[a[j]];
    }
    swap(cur, pre);
  }
  printf("%d\n", add);
  return 0;
}

```