#include <bits/stdc++.h>
const int N = 1000010;
const int moder = (int)1e9 + 7;
int dp[2][N];
int a[N], c[N];
int cnt[N], precnt[N];
int pre[N], tmp[N], pretmp[N];
void add(int &a, int b) {
  a += b;
  a -= a >= moder ? moder : 0;
}
void multadd(int &a, int b, int c) { a = (a + 1ll * b * c) % moder; }
int main() {
  int n, k;
  long long l;
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    c[i] = a[i];
  }
  std::sort(c, c + n);
  int cnt = std::unique(c, c + n) - c;
  for (int i = 0; i < n; ++i) {
    a[i] = std::lower_bound(c, c + cnt, a[i]) - c;
    ++::cnt[a[i]];
  }
  for (int i = 0; i < l % n; ++i) {
    ++tmp[a[i]];
  }
  for (int i = 0; i < N; ++i) {
    precnt[i] = i ? precnt[i - 1] + ::cnt[i] : ::cnt[0];
    pretmp[i] = i ? pretmp[i - 1] + tmp[i] : tmp[0];
  }
  if (l < n) {
    printf("%lld\n", l);
    return 0;
  }
  for (int i = 0; i < N; ++i) {
    dp[0][i] = ::cnt[i];
  }
  int now = 0;
  long long total = l / n;
  int ans = 1ll * n * (total % moder) % moder;
  add(ans, l % n);
  for (int i = 2; i <= k; ++i) {
    for (int j = 0; j < cnt; ++j) {
      pre[j] = dp[now][j];
      if (j) {
        add(pre[j], pre[j - 1]);
      }
    }
    int sum = 0;
    for (int j = 0; j < cnt; ++j) {
      dp[now ^ 1][j] = 1ll * ::cnt[j] * pre[j] % moder;
      if (i - 1 <= total) {
        add(ans, 1ll * tmp[j] * pre[j] % moder);
      }
      add(sum, dp[now ^ 1][j]);
    }
    long long x = total - i + 1;
    if (x > 0) {
      multadd(ans, sum, x % moder);
    }
    now ^= 1;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
const int N = 1000010;
const int moder = (int)1e9 + 7;
int dp[2][N];
int a[N], c[N];
int cnt[N], precnt[N];
int pre[N], tmp[N], pretmp[N];
void add(int &a, int b) {
  a += b;
  a -= a >= moder ? moder : 0;
}
void multadd(int &a, int b, int c) { a = (a + 1ll * b * c) % moder; }
int main() {
  int n, k;
  long long l;
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    c[i] = a[i];
  }
  std::sort(c, c + n);
  int cnt = std::unique(c, c + n) - c;
  for (int i = 0; i < n; ++i) {
    a[i] = std::lower_bound(c, c + cnt, a[i]) - c;
    ++::cnt[a[i]];
  }
  for (int i = 0; i < l % n; ++i) {
    ++tmp[a[i]];
  }
  for (int i = 0; i < N; ++i) {
    precnt[i] = i ? precnt[i - 1] + ::cnt[i] : ::cnt[0];
    pretmp[i] = i ? pretmp[i - 1] + tmp[i] : tmp[0];
  }
  if (l < n) {
    printf("%lld\n", l);
    return 0;
  }
  for (int i = 0; i < N; ++i) {
    dp[0][i] = ::cnt[i];
  }
  int now = 0;
  long long total = l / n;
  int ans = 1ll * n * (total % moder) % moder;
  add(ans, l % n);
  for (int i = 2; i <= k; ++i) {
    for (int j = 0; j < cnt; ++j) {
      pre[j] = dp[now][j];
      if (j) {
        add(pre[j], pre[j - 1]);
      }
    }
    int sum = 0;
    for (int j = 0; j < cnt; ++j) {
      dp[now ^ 1][j] = 1ll * ::cnt[j] * pre[j] % moder;
      if (i - 1 <= total) {
        add(ans, 1ll * tmp[j] * pre[j] % moder);
      }
      add(sum, dp[now ^ 1][j]);
    }
    long long x = total - i + 1;
    if (x > 0) {
      multadd(ans, sum, x % moder);
    }
    now ^= 1;
  }
  printf("%d\n", ans);
  return 0;
}

```