Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

The problem statement has recently been changed. View the changes.

×
C. Riverside Curio

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputArkady decides to observe a river for *n* consecutive days. The river's water level on each day is equal to some real value.

Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the *i*-th day this value is equal to *m*_{i}.

Define *d*_{i} as the number of marks strictly under the water level on the *i*-th day. You are to find out the minimum possible sum of *d*_{i} over all days. There are no marks on the channel before the first day.

Input

The first line contains a single positive integer *n* (1 ≤ *n* ≤ 10^{5}) — the number of days.

The second line contains *n* space-separated integers *m*_{1}, *m*_{2}, ..., *m*_{n} (0 ≤ *m*_{i} < *i*) — the number of marks strictly above the water on each day.

Output

Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.

Examples

Input

6

0 1 0 3 0 2

Output

6

Input

5

0 1 2 1 2

Output

1

Input

5

0 1 1 2 2

Output

0

Note

In the first example, the following figure shows an optimal case.

Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.

In the second example, the following figure shows an optimal case.

Codeforces (c) Copyright 2010-2021 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Aug/03/2021 05:24:51 (i1).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|