order dropdown values highest at the top

I have a set of dropdowns being generate by jquery, but at the moment the lowest value comes out first when I need the opposite, I need in my case the value 12 to always be the one in view.

var selects = [{
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}, {
"12": "Monthly",
"6": "Bi-Monthly",
"4": "Quarterly",
"0": "N/A"
}];
$(function () {
$(".addRow").click(addRow)
})
function sum($els) {
var sum = 0;
$els.each(function () {
sum += +$(this).val()
})
return sum
}
function addRow() {
var fieldset = $('<fieldset></fieldset>').appendTo("#test");
var sumDiv = $("<div class='sum'>total £<span>0</span></div>").appendTo(fieldset)
$.each(selects, function (i, v) {
var select = $("<select name='select" + i + "'></select>").prependTo(fieldset)
$.each(v, function (value, key) {
$("<option></option>", {
value: value,
text: key
}).appendTo(select)
})
select.change(function () {
fieldset.find("span").text(sum($("select", fieldset)))
$(".sumTotal").find("span").text(sum($("#test select")))
}).change()
})
}

 

Two things, if you want the list to be as you have in the array, switch:

appendTo(select)

Code (markup):

To

prependTo(select)

Code (markup):

If you want the first item selected then, you would do at the end of the addRow function:

select[0].selectedIndex = 0;

Code (markup):