#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
long long n, l, k;
vector<vector<int> > dp;
vector<int> sdp;
long long a[1 << 20];
pair<long long, int> sa[1 << 20];
int nn[1 << 20];
int main() {
  scanf("%lld %lld %lld", &n, &l, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
    sa[i] = make_pair(a[i], i);
  }
  sort(sa, sa + n);
  for (int i = 0; i < n;) {
    int no = i;
    while (no < n && sa[i].first == sa[no].first) {
      ++no;
    }
    for (int j = i; j < no; ++j) {
      nn[sa[j].second] = no - 1;
    }
    i = no;
  }
  long long ans = 0;
  dp.assign(k, vector<int>(n));
  sdp.resize(n);
  for (long long i = 0; i < n; ++i) {
    dp[0][i] = 1;
  }
  for (int i = 0; i < n; ++i) {
    if (i == 0) {
      sdp[i] = dp[0][sa[i].second];
    } else {
      sdp[i] = sdp[i - 1] + dp[0][sa[i].second];
    }
    if (sdp[i] >= mod) {
      sdp[i] %= mod;
    }
  }
  for (long long j = 1; j < k; ++j) {
    for (long long i = 0; i < n; ++i) {
      dp[j][i] = sdp[nn[i]];
    }
    for (int i = 0; i < n; ++i) {
      if (i == 0) {
        sdp[i] = dp[j][sa[i].second];
      } else {
        sdp[i] = sdp[i - 1] + dp[j][sa[i].second];
      }
      if (sdp[i] >= mod) {
        sdp[i] %= mod;
      }
    }
  }
  ans = 0;
  for (int j = 0; j < k; ++j) {
    for (int i = 0; i < n; ++i) {
      long long mul = l / n;
      if (i < l % n) {
        ++mul;
      }
      mul -= j;
      if (mul > 0) {
        if (mul >= mod) {
          mul %= mod;
        }
        ans += (long long)dp[j][i] * mul;
      }
      if (ans >= mod) {
        ans %= mod;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int mod = 1000000007;
long long n, l, k;
vector<vector<int> > dp;
vector<int> sdp;
long long a[1 << 20];
pair<long long, int> sa[1 << 20];
int nn[1 << 20];
int main() {
  scanf("%lld %lld %lld", &n, &l, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
    sa[i] = make_pair(a[i], i);
  }
  sort(sa, sa + n);
  for (int i = 0; i < n;) {
    int no = i;
    while (no < n && sa[i].first == sa[no].first) {
      ++no;
    }
    for (int j = i; j < no; ++j) {
      nn[sa[j].second] = no - 1;
    }
    i = no;
  }
  long long ans = 0;
  dp.assign(k, vector<int>(n));
  sdp.resize(n);
  for (long long i = 0; i < n; ++i) {
    dp[0][i] = 1;
  }
  for (int i = 0; i < n; ++i) {
    if (i == 0) {
      sdp[i] = dp[0][sa[i].second];
    } else {
      sdp[i] = sdp[i - 1] + dp[0][sa[i].second];
    }
    if (sdp[i] >= mod) {
      sdp[i] %= mod;
    }
  }
  for (long long j = 1; j < k; ++j) {
    for (long long i = 0; i < n; ++i) {
      dp[j][i] = sdp[nn[i]];
    }
    for (int i = 0; i < n; ++i) {
      if (i == 0) {
        sdp[i] = dp[j][sa[i].second];
      } else {
        sdp[i] = sdp[i - 1] + dp[j][sa[i].second];
      }
      if (sdp[i] >= mod) {
        sdp[i] %= mod;
      }
    }
  }
  ans = 0;
  for (int j = 0; j < k; ++j) {
    for (int i = 0; i < n; ++i) {
      long long mul = l / n;
      if (i < l % n) {
        ++mul;
      }
      mul -= j;
      if (mul > 0) {
        if (mul >= mod) {
          mul %= mod;
        }
        ans += (long long)dp[j][i] * mul;
      }
      if (ans >= mod) {
        ans %= mod;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```