树状数组+dp

B - Sereja and Subsequences

Sereja has a sequence that consists of n positive integers, a1, a2, ..., an.

First Sereja took a piece of squared paper and wrote all distinct non-empty non-decreasing subsequences of sequence a. Then for each sequence written on the squared paper, Sereja wrote on a piece of lines paper all sequences that do not exceed it.

A sequence of positive integers x = x1, x2, ..., xr doesn't exceed a sequence of positive integers y = y1, y2, ..., yr, if the following inequation holds: x1 ≤ y1, x2 ≤ y2, ..., xr ≤ yr.

Now Sereja wonders, how many sequences are written on the lines piece of paper. Help Sereja, find the required quantity modulo 1000000007 (109 + 7).

Input

The first line contains integer n (1 ≤ n ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106).

Output

In the single line print the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

1
42

Output

42

Input

3
1 2 2

Output

13

Input

5
1 2 3 4 5

Output

719

题意：给你一个序列，要你写出它的所有子序列，并求它的所有子序列的价值和，所谓价值和就是该子序列所有元素的乘积

分析：设dp[i]表示以位置i结尾的子序列的最大价值和，可以递推出，dp[i]=前i-1的所有子序列的价值和乘以a[i]+单独以a[i]为一个子序列的价值（就是a[i]），还要注意要去重，如果子序列含有重复的数，要把以它为结尾的子序列的最大价值和减掉

AC code:
 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn=1e6+5;
const ll mod=1e9+7;
ll c[maxn];
ll dp[maxn];
ll a[maxn];
ll pre[maxn];
ll n;
ll lowbit(ll x)
{
    return x&(-x);
}
ll getsum(ll x)
{
    ll ans=0;
    while(x)
    {
        ans+=c[x];
        ans%=mod;
        x-=lowbit(x);
    }
    return ans%mod;
}
void update(ll x,ll val)
{
    while(x<=1000010)///很重要，不是n了，传进来的是a[i]
    {
        c[x]+=val;
        c[x]%=mod;
        x+=lowbit(x);
    }
}
int main()
{
    ios::sync_with_stdio(0),cin.tie(0);
    cin>>n;
    memset(dp,0,sizeof dp);
    memset(pre,-1,sizeof pre);
    for(int i=1;i<=n;i++)
        cin>>a[i];
    ll ans=0;
    for(int i=1;i<=n;i++){
        dp[i]=getsum(a[i])*a[i]+a[i];///状态转移dp[i]表示以i结尾的子序列能达到的最大价值和
                                     ///所以dp[i]=前i-1的子序列的总价值和再乘上第i个子序列的价值+第i个数
                                     ///单独作为一个子序列
        if(pre[a[i]]!=-1) dp[i]-=pre[a[i]];///减去重复的
        dp[i]=(dp[i]%mod+mod)%mod;///防止dp[i]减重而小于0
        ans+=dp[i];
        ans=(ans%mod+mod)%mod;
        pre[a[i]]=getsum(a[i])*a[i]+a[i];
        update(a[i],dp[i]);///每次更新第a[i]位置的值为以i结尾的子序列能达到的最大价值和
    }
    cout<<(ans%mod+mod)%mod<<endl;
    return 0;
}