Codeforces 314C. Sereja and Subsequences，codeforces314c

Codeforces 314C. Sereja and Subsequences，codeforces314c

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Codeforces JAVA8 比 JAVA6 慢了快一倍.......

C. Sereja and Subsequencestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

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 modulo1000000007 (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).

Sample test(s)input

142

output

42

input

31 2 2

output

13

input

51 2 3 4 5

output

719

/** * Created by ckboss on 14-9-8. */import java.util.*;public class SerejaandSubsequences {    static final long mod = 1000000007;    static int n;    static long[] a = new long[100100];    static long[] tree = new long[1000100];    static long[] dp = new long[100100];    static int lowbit(int x){        return x&(-x);    }    static void ADD(int pos,long value){        for(int i=pos;i<=1000010;i+=lowbit(i)){            tree[i]+=value;            if(tree[i]>=mod) tree[i]-=mod;        }    }    static long SUM(int pos){        long ret=0;        for(int i=pos;i>0;i-=lowbit(i)){            ret+=tree[i];            if(ret>=mod) ret-=mod;        }        return ret;    }    public static void main(String[] args){        Scanner in = new Scanner(System.in);        n=in.nextInt();        for(int i=1;i<=n;i++){            a[i]=in.nextInt();            long sum=SUM((int)a[i]);            long jian=SUM((int)a[i])-SUM((int)(a[i]-1));            long temp=((sum*a[i])%mod+a[i]-jian+mod)%mod;            dp[i]=(dp[i]+temp)%mod;            ADD((int)a[i],temp);        }        long ans=0;        for(int i=1;i<=n;i++){            ans=(ans+dp[i])%mod;        }        System.out.println(ans);    }}